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Question: Saponification of ethyl acetate by NaOH is a second order reaction with rate constant K \( = \,6.36\...

Saponification of ethyl acetate by NaOH is a second order reaction with rate constant K = 6.36 litre mol−1 min⁡−1 = \,6.36\,litre\,mo{l^{ - 1}}\,{\min ^{ - 1}}at 25∘C{25^\circ }C.
The initial rate of the reaction when the base and the ester have concentration = 0.02 mol litre−1? = \,0.02\,mol\,litr{e^{ - 1}}?
What will be the rate of reaction 1010minutes after the commencement of the reaction??
A) 2.544×10−3 mol litre−1 min⁡−12.544 \times {10^{ - 3}}\,mol\,litr{e^{ - 1}}\,{\min ^{ - 1}} (b) 4.925×10−4 mol litre−1 min⁡−1 \left( b \right)\,4.925 \times {10^{ - 4\,}}mol\,litr{e^{ - 1}}\,{\min ^{ - 1\,}}
B) 2.544×10−4 mol litre−1 min⁡−1 (b) 49.25×10−4 mol litre−1  min⁡−1\,2.544 \times {10^{ - 4}}\,mol\,litr{e^{ - 1}}\,{\min ^{ - 1}}\,\left( b \right)\,49.25 \times {10^{ - 4}}\,mol\,litr{e^{ - 1\,}}\,{\min ^{ - 1}}
C)  25.44×10−3 mol litre−1 min⁡−1 (b) 4.925×10−5 mol litre−1 min⁡−1\,25.44 \times {10^{ - 3}}\,mol\,litr{e^{ - 1}}\,{\min ^{ - 1}}\,\left( b \right)\,4.925 \times {10^{ - 5}}\,mol\,litr{e^{ - 1}}\,{\min ^{ - 1}}
D) None of these

Explanation

Solution

‌The reaction is a second order reaction thus we can use differential rate law for second order reactions to calculate initial rate and rate after time t and also, we can use integrated rate law expression for second order reactions to calculate concentration of the reactants after time t.As clearly provided in the question, second order rate kinetics is involved in this question.

Complete step by step solution:
So, the Initial rate using differential rate law expression is
r0=k[A]0[B]0{r_0} = k{\left[ A \right]_0}{\left[ B \right]_0} 6.36 litre mol−1min⁡−1×0.02 mol litre−1×0.02 mol litre−1 =2.544×10−3mol litre−1 min⁡−16.36\,litre\,mo{l^{ - 1}}{\min ^{ - 1}} \times 0.02\,mol\,litr{e^{ - 1}} \times 0.02\,mol\,litr{e^{ - 1\,}} = 2.544 \times {10^{ - 3}}mol\,litr{e^{ - 1\,}}{\min ^{ - 1}}
Now, Concentration of reactants after time t=10 minutes is calculated using integrated rate law expression is 1At=1A0+kt\dfrac{1}{{{A_t}}} = \dfrac{1}{{{A_0}}} + kt
1At=10.02+(6.36×10) litre mol−1\dfrac{1}{{{A_t}}} = \dfrac{1}{{0.02}} + \left( {6.36 \times 10} \right)\,litre\,mo{l^{ - 1}}
1At=50+63.6\dfrac{1}{{{A_t}}} = 50 + 63.6
At=1113.6 mol litre−1{A_t} = \dfrac{1}{{113.6}}\,mol\,litr{e^{ - 1}}
Concentration of reactants after time t=10 minutes =1113.6mol litre−1\dfrac{1}{{113.6}}mol\,litr{e^{ - 1}} =0.0088 mol litre−1 = 0.0088\,mol\,litr{e^{ - 1}}
Also, Rate after time t=10 minutes
r=k[A][B]r = k\left[ A \right]\left[ B \right]
6.36×0.0088×0.0088=4.925×10−4  mol litre−1min⁡−16.36 \times 0.0088 \times 0.0088 = 4.925 \times {10^{ - 4\,}}\,mol\,litr{e^{ - 1}}{\min ^{ - 1}}

The correct option is Option A.

Additional Information: For a second order reaction,
A +B→ PA\, + B\xrightarrow{{}}\,P
The differential rate equation is given as follows:
Rate=−d[A]dt=−d[B]dt=d[P]dt=k[A][B]Rate = \dfrac{{ - d\left[ A \right]}}{{dt}} = \dfrac{{ - d\left[ B \right]}}{{dt}} = \dfrac{{d\left[ P \right]}}{{dt}} = k\left[ A \right]\left[ B \right]
On, integrating the above equation, we get
1At=1A0+kt\frac{1}{{{A_t}}} = \dfrac{1}{{{A_0}}} + kt

Note: We use the differential rate law expression and integrated rate law expression to calculate rate of the reaction and concentration of the reactants after time t.The overall order of the reaction is 2, first order with respect to A(NaOH) and first order with respect to B(ester) so the total order of the reaction is 2.